Let $\Bbb E^n$ be the Euclidean $n$-space. Let $p \in \Bbb S^{n-1},$ the $(n-1)$-sphere in $\Bbb E^n.$ Let $w \in T_p \Bbb S^{n-1},$ the subspace of $\Bbb E^n$ parallel to the affine subspace tangent to $\Bbb S^{n-1}$ at $p.$ Consider the map $\gamma_w : (-1,1) \longrightarrow \Bbb E^n$ defined by $$\gamma_w (t) = \frac {p+wt} {\sqrt {\left \langle p +wt, p+ wt \right \rangle}},\ t \in (-1,1).$$ Prove that $\dot {\gamma_w} (0) = w.$
Consider the map $g : \Bbb E^n \longrightarrow \Bbb E^n$ defined by $g(v) = \frac {v} {\sqrt {\left \langle v, v \right \rangle}},\ v \in \Bbb E^n.$ and the map $f : (-1,1) \longrightarrow \Bbb E^n$ defined by $t \mapsto p+tw,\ t \in (-1,1).$ Then $\gamma_w = g \circ f.$ So by Chain rule we have $$\dot {\gamma_w} (0) = D\gamma_w (0) (1) = Dg(f(0)) (w) = Dg(p) (w).$$
How do I compute $Dg(p)\ $? Can anybody please help me in this regard?
Thanks for your time.
The differential $Dg$ is just given by the Jacobian matrix. You can compute it in coordinates easily. If $v = (x_1,\dots,x_n)$, then
$$ g(v) = (g_1,\dots,g_n) = \left( \frac{x_1}{\sqrt{x_1^2+\cdots+x_n^2}}, \; \dots, \; \frac{x_n}{\sqrt{x_1^2+\cdots+x_n^2}} \right) $$
You can compute the partial derivates using the quotient rule. You get $$ \frac{\partial g_i}{\partial x_i} = \frac{|v|^2-x_i^2}{|v|^3} \quad \text{and} \quad \frac{\partial g_i}{\partial x_j} = \frac{-x_ix_j}{|v|^3} $$ where $i \neq j$. Putting it all together, you get $$ Dg(v) = \frac{1}{|v|^3}\begin{pmatrix} |v|^2-x_1^2 & -x_1x_2 & \cdots & -x_1x_n \\ -x_2x_1 & |v|^2-x_2^2 & \cdots & -x_2x_n \\ \vdots & \vdots & \ddots & \vdots \\ -x_nx_1 & -x_nx_2 & \cdots & |v|^2-x_n^2 \end{pmatrix} = \frac{1}{|v|} \mathrm{Id}_n - \frac{1}{|v|^3} v \, v^\top $$ Here, $v$ is a column vector, and $v^\top$ is the transpose row vector, so $v \, v^\top$ is the matrix whose entries are $x_ix_j$.
When $p$ is on the unit sphere, $|v|=1$, and this simplifies to: $$ Dg(p) = \mathrm{Id}_n - p \, p^\top $$
Now plug in $w$. The first term gives $\mathrm{Id}_n w = w$, and since $w$ is in the tangent plane to $p$, you have $p^\top w = 0$, so the second term vanishes.